Difference between revisions of "2023 AMC 12B Problems/Problem 23"
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+ | ==Solution== | ||
+ | The product can be written as | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2^a 3^b 4^c 5^d 6^e | ||
+ | & = 2^{a + 2c + e} 3^{b + e} 5^d . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>. | ||
+ | We denote this number as <math>f(n)</math>. | ||
+ | |||
+ | Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>. | ||
+ | |||
+ | Thus, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | f \left( n \right) | ||
+ | & = \sum_{d = 0}^n g \left( n - d \right) \ | ||
+ | & = \sum_{k = 0}^n g \left( k \right) . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Next, we compute <math>g \left( k \right)</math>. | ||
+ | |||
+ | Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>. | ||
+ | Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | g \left( k \right) | ||
+ | & = \sum_{i=0}^k \left( 2 k - i + 1 \right) \ | ||
+ | & = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | f \left( n \right) | ||
+ | & = \sum_{k = 0}^n g \left( k \right) \ | ||
+ | & = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \ | ||
+ | & = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2 | ||
+ | - \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \ | ||
+ | & = \frac{3}{2} \cdot | ||
+ | \frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right) | ||
+ | - \frac{1}{2} \cdot | ||
+ | \frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \ | ||
+ | & = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | By solving <math>f \left( n \right) = 936</math>, we get<math>n = \boxed{\textbf{(A) 11}}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 17:41, 15 November 2023
Solution
The product can be written as
Therefore, we need to find the number of ordered tuples where , , , , are non-negative integers satisfying . We denote this number as .
Denote by the number of ordered tuples where with .
Thus,
Next, we compute .
Denote . Thus, for each given , the range of is from 0 to . Thus, the number of is
Therefore,
By solving , we get.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)